 Hello everyone, I am Mr. D. J. Doshi, Assistant Professor, Department of Mechanical Engineering, Walsh Institute of Technology, SULOP. In this video, we will be studying about the problem related to projection of solids. At the end of this session, students will be able to draw the projections of the solids as per given conditions. Now, this is called a tetrahedron. So what is tetrahedron? First, so tetrahedron is a solid of which all the four faces are equilateral triangle. So this is nothing but a pyramid of triangular base that is triangular base pyramid or triangular pyramid of which all the faces that is 3 plus 1, four faces will be equilateral triangles. So let us read the problem, a tetrahedron of 45 mm side has one of its edge in HP and inclined at 45 degree to AP. The triangular place contained by the edge of a base in the HP is perpendicular to the HP, draw its projections. So one by one we will start, first thing it is a tetrahedron of 45 mm side base. So all the sides will be of 45 mm. One of its edge is in HP, so edge is in HP. If you observe the tetrahedron, its top view will be of course a equilateral triangle with apex O joined with 1, 2, 3 that is 1, 2, 3 is a triangle which is equilateral triangle of 45 mm side and O1, O2 and O3 will be the generators which will be shown by the dark line. Now the condition is that its base is in HP. So now what is the base edge 1, 2, 3 and 3, 1 are the base edges, out of that one edge should be in HP. So you have to draw the equilateral triangle in such a way that 2, 3 will be towards your right hand side and it will be perpendicular to VP and parallel to HP or resting on HP. So draw equilateral triangle in the top view in such a way that 1, 2, 3 is equilateral triangle and 2, 3 will be perpendicular to VP. Now project upwards projected 1 upwards, so give it as a 1 dash and project 2, 3 upwards so it will be 2 dash, 3 dash, O is apex which is at the center of the equilateral triangle. Now projected upwards you will get O, O dash, so extend it upwards. Now see O1, O2 and O3 are the generators of the tetrahedral, out of that if you observe O1 is parallel to XY line or parallel to VP. So now as O1 is parallel its front view projected upwards will be a two length line. As we know the basic thing that if one of the view is parallel to one of the plane it will be viewed in another plane as a two length line. So O1 in the top view is parallel to XY line or parallel to the plane. So project 1 upwards and O upwards and O upwards. Now 1 O will be 45 degrees, so take 1 as a center and with compass and length equal to two length line that is 45 mm mark the apex as 45 degree on the projection of O. So you will get the O dash point, so O dash 1 dash will be equal to 45 degree and join O dash 2 dash and O dash 3 dash, of course 2 dash 3 dash will be a point view of a line 2 3 that is base H which is resting on HP. So you will get a triangle of which O dash 1 dash will be two length line, O dash O dash 2 dash and O dash 3 dash which is the same line is a apparent length of the generator O2 and O3. Now complete this projection later on what is given a tetrahetron of 45 mm side has one of its H in HP and inclined at 45 degree VP. So VP we will be drawing later on the angle with VP we will be drawing in the third stage. Now what is next step given that the base H containing that H in HP is perpendicular to HP. So now what is the base H joining to apex is O2 or O3 dash O2 O dash 2 dash or O3 dash O dash 3 dash which is a line, so it should be perpendicular to HP, so we will rotate this front view triangle in such a way that the line O dash 2 dash or O dash 3 dash which is the same line perpendicular to HP. So rotate now see if you observe this is the view and this is the phase O dash 2 dash 3 dash and this has to be perpendicular to HP. So I have rotated in clockwise direction in such a way that O dash 2 dash 3 dash which is a line view or a edge view of a triangle. So it will be perpendicular to HP and redraw this. So first draw O dash 2 dash 3 dash mark O dash from there mark take the distance 45 mm that is O dash 1 dash and take the base distance with the compass at 2 dash 3 dash mark here join 1 dash 2 dash 3 dash and 1 dash O dash and of course the axis will be drawn here. So now from this we will observe that the triangular place phase contained by the edge of the base in the HP is perpendicular to HP. So now this is the phase O2 O dash 2 dash 3 dash which is containing the base edge 2 dash 3 dash is perpendicular to HP and we have redrawn this. Now projected upwards so O dash 2 dash and 3 dash are on the same projector projected downer project 3 horizontally you will get 0.3 project O horizontally you will get 0.0 here and project 2 and vertically 2 you will get 0.2 now here we have 1. So 1 we will project it downwards 1 projected horizontally you will get 0.1. So join 1 3 1 2 and 1 O. So this is the top view of the second stage when the phase containing that base edge is perpendicular to HP. So this is the phase containing that edge that is base edge is perpendicular to HP. Now next part that is inclined to 45 degree to HP. So now as this is the point view of line 2 3 so this will be of course a 2 length line we have already drawn the equilateral triangle in the first stage same is projected here. So 2 O dash 3 will be a 2 length line that is edge view of the base you can say. So now what is given it is inclined at 45 degree to HP. So draw a 45 degree line on this making 45 degree with x y line on that take certain point 3 away from the plane and join 3 O 2 which is of course of the same length and redraw the triangle 3 1 2 3 and mark 1 O dash 1 O. So this is the plane top view of the plane making rotated in such a way that the base edge 3 base edge or edge view of the base is making 45 degree with VP. So this is a 45 degree line on which we have completed the top view of the second stage. Now project 1 upwards project horizontally 1 dash you will get 0.1 project 3 upwards horizontally of course 3 will be on the base and project 2 upwards you will get 2 dash which is of course on the again x y line on HP and project O dash here O dash here O you will get O dash. Now if you observe 0.3 is away from the observer because the observer is observing from this side 0.3 is away from the observer. So the lines related to 3 will be dotted lines. So if you observe here 3 O 3 2 and 3 1 are the dotted should be the dotted lines. So if you observe here 3 2 or 2 dash 3 dash is the base on which it is resting. So you cannot draw 3 dash 2 dash as a dotted line again 3 1 though it is not visible from here but 3 1 is a peripheral line of the tetrahedral view. So you also cannot draw the line 3 1 as a dotted line. So what we have understood from here was the video and think over it see though we need to draw 3 1 and 3 2 as a dotted line we are not drawing y. So it is clear from here that the peripheral lines must not be dotted though those are not visible. So we will not be drawing 3 1 and 3 2 as dotted line but 3 1 3 O can be a dotted line. So project 3 upwards as I told you 3 2 1 so 1 dash 3 dash 2 dash you will get. So join the points 1 dash 2 dash as a dark line because it is a visible from here. Join 1 dash 3 dash though it must have been dotted line but as it is a peripheral line we will draw this as a dark line and 2 dash 3 dash is again a peripheral line which is resting on HP. So 2 dash 3 dash also will be a dark line. So 1 dash 2 dash 2 dash 3 dash and 3 dash 1 dash will be dark lines. Now join 1 O of course that is visible clear and 2 O that is 2 dash O of course is a visible line. Only the line is a dotted line is O dash 3 dash. So complete the view and this will give you the final view where the tetrahedral 45 sides and the face is making perpendicular to HP and the base is inclined at 45 degree to VP. So here we have made the face perpendicular to HP and here we have made base 45 degree to VP. As this is a point view this will be a two length line. So this also will be a two length line. So this also will be a two length line. So need not to go for apparent length or two length here. So it will be of course a two length line projected upwards and complete. For this we have used an engineering drawing by P. H. Jain. Thank you.